Find $\int \dfrac{1}{3x^2+6x+78}\,dx$. Choose 1 answer: Choose 1 answer: (Choice A) A $\text{arcsin}\left(\dfrac{x+1}{5}\right)+C$ (Choice B) B $ \text{arctan}\left(\dfrac{x+1}{5}\right)+C$ (Choice C) C $\dfrac{1}{15} \text{arcsin}\left(\dfrac{x+1}{5}\right)+C$ (Choice D) D $\dfrac{1}{15} \text{arctan}\left(\dfrac{x+1}{5}\right)+C$
Solution: The integrand is in the form $\dfrac{1}{p(x)}$ where $p(x)$ is a quadratic expression. This suggests that we should rewrite $p(x)$ by completing the square. Specifically, we will rewrite $p(x)$ as a constant multiple of $(x+h)^2+k^2$. Then, we will be able to integrate using the following formula, which is based on the derivative of the inverse tangent function: $\int \dfrac{1}{(x+ h)^2+ k^2}\,dx=\dfrac{1}{ k} \text{arctan}\left(\dfrac{x+ h}{ k}\right)+C$ [Why is this formula true?] We start by rewriting $p(x)$ as a constant multiple of $(x+ h)^2+ k^2$ : $\begin{aligned} 3x^2+6x+78&=3[x^2+2x+26] \\\\ &=3[x^2+2x+1+25] \\\\ &=3[(x+1)^2+25] \\\\ &=3[(x+{1})^2+{5}^2] \end{aligned}$ Now we can find the integral: $\begin{aligned} &\phantom{=}\int \dfrac{1}{3x^2+6x+78}\,dx \\\\ &=\int\dfrac{1}{3[(x+1)^2+5^2]}\,dx \\\\ &=\dfrac13\int\dfrac{1}{(x+{1})^2+{5}^2}\,dx \\\\ &=\dfrac13\cdot\dfrac{1}{{5}} \text{arctan}\left(\dfrac{x+{1}}{{5}}\right)+C \\\\ &=\dfrac{1}{15} \text{arctan}\left(\dfrac{x+1}{5}\right)+C \end{aligned}$ In conclusion, $\int \dfrac{1}{3x^2+6x+78}\,dx=\dfrac{1}{15} \text{arctan}\left(\dfrac{x+1}{5}\right)+C$